The errors vary 0-2 degrees, ie not much at all.

Assume that intermediate headings have an error varying as a straight line (you have no basis for presuming anything else) between the two bounding stated headings and just interpolate linearly (ie as a straight line).

So, as an example

090M = 090C (nil error) 120M = 122C (compass reads 2 degrees more than magnetic)

If we wanted to know an intermediate reading, we would observe that the error varies from 0 (090M) to +2 (120M). So, if we presume a linear (straight line) variation, that would be 2 degrees error variation for the 30 degrees (090 to 120 range). If we wanted an idea of the compass heading for, say, 105M (midpoint in the range 090 to 120), we would put that at 1 degree error and conclude that 105M would be pretty close to 106C.

If you come up with a fraction of a degree, just round it off to the nearer whole degree in the usual arithmetic process that 0.5 and above rounds to the higher value while 0.4999 (etc) rounds to the whole lower value. So, for example, if we wanted the compass heading for 100M, that would be 0 + 1/3 of 2 = 0.667 so we would call it 1 degree error and round the compass heading off to 101C.

... which is what Bob says but in a slightly different way. If it still doesn't make sense, please do say so and we'll keep varying the story slightly until it does.]]>

35....206....41....158....32....120 is correct

I have amended the figure on Page 59 as well

Cheers]]>

Best of Luck.]]>

Your correct Radio waves spread out in a straight line in all directions from the source and can only follow a great circle path.

(or take the shortest path) There is a question about it Progress test 1 on page 31.]]>

Ravi Mahajani]]>

Ravi]]>

As long as you remain below 6000ft until 50nm, it doesn't matter what happens after that because the controlled airspace steps up to 8000ft after 50nm, so the last part of the climb from 6000ft to 6500ft can be ignored.

All that matters is to maintain your present level until you can climb through 3500ft (6000 - 2500) by 50nm out. That takes 7min @ 500ft/min and that's 11.7nm @ 100kt.

The question asked for the distance from A to commence the climb, so that's 50 - 11.7 = 38.3nm from A.]]>

As noted in the last post, it moves all over the shop, (generally) in a small sort of way, and the rate varies significantly with time. Hence the background effort to keep a close eye on the position and refine predicted movement in the short term.

There are plenty of useful links on the net on the subject so you can read away to your heart's content. One, at random, www.ncei.noaa.gov/news/tracking-changes-earth-magnetic-poles, should provide some time over a coffee or two.]]>

Presuming that you are looking at a constant ROD/ROC, there will be a range of values which is feasible for your particular aircraft. If the gradient required is beyond the capability of your aircraft, then you will need to orbit for part of the climb/descent to get around that problem. If you end up flying airlines, you will find this problem to be a common requirement as ATC does its aircraft separation juggling act.

In earlier days, we just did the sums in our heads as mental arithmetic - these days, the FMC/FMS kits will do the work for you, if you want to go that way. One of the difficulties, these days, is that many pilots have a real problem using the FMC/FMS as a mental crutch.

A sideline problem is where the aircraft performance outstrips the IVSI’s range. For example, on the B727, it was very common to have the IVSI pegged on the stop during a descent. In this case, the pilot has to run continuous calculations on the way down, looking at altimeter changes either in a time or distance interval, depending on just what you were trying to achieve. Some civil aircraft can come down quite quickly - the IAI Westwind is a good example.

It gets more interesting in the fast jet community. I had a mate relate a tale from his F/A 18 days – given an unrestricted climb to a high level, ATC required that he advise leaving each 1000 foot level. Sitting the aircraft on its tail, his transmission was along the lines of “left 1000 2000 3000 4000 .. " finishing at TOC. As he observed, the ATC-er concerned learned a simple lesson regarding fast jet performance on that occasion.

The easiest way to map values to the CTA/CTR profile is to start with a known spot - either where you need to start or finish the climb/descent and then work out the gradient (ft/nm) from that point to each of the step vertices in turn. This should cover any real world situation you might encounter.

From the resulting gradients, it becomes evident which one is limiting. That's the step which you need to watch like a hawk all the way up or down. On occasion, you may have multiple steps which give very similar gradients and you need to monitor more than one.

If that doesn’t give you enough information, say so and I’ll run up some graphics to amplify the comments.]]>

Now, there are three ways to go about the question -

(a) start your climb immediately, as you did, and blast up to your desired cruise level. However, does that create any problems along the way which, perhaps, you may have overlooked ?

(b) delay the start of your climb if, and as, necessary.

(c) start your climb immediately, with a ROC to suit any, and all, constraints.

Normally, one would run with (b).

Now, getting back to (a), what trap did you fall into ? Let me give you a clue -

The distance covered between the two would be 50nm-22nm=28nm

might offer a suggestion ....

If you end up in airlines, expect to see this from your check captain on a route check from time to time. Stock standard problem to pose to F/Os in particular.]]>

That's the trick. First, one needs to learn the background stuff. Then it is just practice, practice, practice. The only two things in the exam which present a problem are speed and accuracy.

Did I mention practice, practice, practice ?]]>

When you look at your sketch, you might see something which suggests that you may have omitted an item of data in the original question ?]]>

Again, this just needs a bit of reflection and thought to sort out the answer.

Easiest to work in gradients of ft/nm for comparison, keeping in mind that we will need to "clip" one (or more) steps. Keep in mind that the published height limits are OCTA so we can cruise or clip steps at those heights.

First, your speed is 120 kt or 2 nm/min. At a ROC of 800 ft/min, this is equivalent to a climb gradient of 400 ft/nm.

Looking at the airspace steps -

(a) from your present position to the next step is a 2000 ft climb in 13 nm so the gradient is 2000/13 = 154 ft/nm, well below your climb gradient so you will need to delay the climb so you don't infringe the 4000 ft airspace step.

(b) next step is 4000 ft to 6000 ft in (50-35) = 15 nm for a gradient of 2000/15 = 133 ft/nm, again well below your climb gradient.

(c) once you are past 50 nm, the step is 10000 which is not a concern for an 8500 ft cruise. We now have the overall picture of the climb requirement and you will be planning to just clip 6000 ft at 50 nm.

While we don't expect a problem, just check by working backwards to make sure that no other steps present a concern.

If we start at 6000 ft and 50 nm and work back at 400 ft/nm, then at the 35 nm step, we have just travelled 15 nm for 15 x 400 = 6000 ft. It is pretty obvious that no other steps will present a limitation problem so we just have to work back from the 6000 ft at 50 nm step to figure out where we need to start the climb from 2000 ft. That is to say, how far will it take for us to climb 6000 - 2000 = 4000 ft ? That is easy enough to figure out as 4000 ft / 400 ft/nm = 10 nm (always be on the lookout for easy mental arithmetic - saves lots of time in the exam).

The answer then is that we start the climb 10 nm before the 50 nm step, ie 40 miles from A, so the answer you are after is "D".]]>

If you have a look at the daylight and darkness charts in the AIP, you can figure that flying north or south will approximate a daylight change rate of 1 degree for 2 minutes. Using the 15 degrees = 1 hour change for longitude, we would be looking at 1 degree = 4 minutes. End result is that we get a quicker result if we fly east or west rather than north or south. As we are reasonably near to the equator, the latitude and longitude distances per degree are not too different so, to a first order approximation, we might take them as being much the same for the purposes of this question.

Flying east would be the way to go.]]>

Please find attached image.

I measured from the latest edition of the SYD WAC and the distance off track to Hill end is actually 5nm.

Thanks,

Jason]]>

Thanks Bob]]>

The answer is d and I've amended the book to include the answer]]>

Congratulations on an excellent result. Well done!!]]>

Cheers, Erin]]>

I have a better understanding of them now!! Thank you!!]]>

1. Heading change required

2. Heading required

I believe you'd agree they have a different meaning, where 1. would be the delta, 2. would be the number on a DG.

Hope this helps.]]>

1. calculate drift (what winds have done to you). at this point in time you already know that you are off the track.

2. when you apply drift to your heading, it will bring you to the track

3. to intercept orginal track you need to correct more, otherwise you will end up abeam (parallel) to your destination

For weather avoidance in real life, read paragraph "Other applications" on page 112 Cpl Navigation book (ed. 2015). Basically use same angles (from your DG) and same times (using your watch).

Hope this helps]]>

First, a useful little mnemonic - which appears to date back to the quite early days of US military aviation - "iced tea is a pretty cool drink" to cover the order of things -

is a

gives the sequence IAS- position error - CAS - compressibility correction - EAS - density correction - TAS

The equation to correct EAS to TAS is shown below. As we are concerned with low performance aircraft, we will ignore EAS and use CAS in its place.

TAS = CAS/√σ

where sigma (σ) = density ratio = ρ/ρo

where ρ (rho) is the local density and ρo is the standard sea level density (1.225 kg/m

As an aside, when we play with performance stuff, we make considerable use of non-dimensionalised numbers (ratios, in this case). You may come across references to sigma (as above), delta (δ - ratio of pressures), and theta (Θ - ratio of temperatures) in your general aviation reading.

Now if we know CAS and sigma, for our low performance aircraft flying along at whatever height, then we can figure out the conversion from CAS to TAS. For low speed (ie non-compressible) flight, any of the navigation computers will do this for us.

Bosi, (that nice chap) has posted a scan of a Dalton (E6B) computer with this stuff shown, so we will use his scan for an example.

Now, if we were to look up a table of ISA values (you can get these from the net via Mr Google), we would find that the density at 10,000 ft is 0.7385 kg/m

Looking at Bosi's scan, the airspeed cutout is set to give a density height of 10,000 ft. This setting (known as a nomogram), while providing the density height for us, is not intended solely for that purpose. Its main purpose is to set the outer slide rule C/D scales to specific values.

If you look at the settings Bosi has ringed, at the position where the outside scale is "10" shows, on the inner scale, a value of 0.86. Notice that this is the value we worked out, above, for √σ = 0.8594, rounded off to two decimals for the slide rule precision.

Similarly, if you look at the position where the inner scale is "10", this shows, on the outer scale, 1.161. Again, notice that this is the value we worked out, above, for 1/√σ = 1.1637. The discrepancy simply reflects the fact that we can't read the third decimal all that accurately on the slide rule.

So, what the airspeed cutout is doing, is setting up the multiplications

TAS = CAS/√σ , and

CAS = TAS x √σ

depending on which way you want to do things - same result, though.

So, we can use the whizz wheel to get the value of sigma for any height and then multiple this by the standard sea level pressure to get the density at that height. You can use this to impress your colleagues at the Aero Club bar on a Saturday afternoon, should you so desire ....

Just a couple of clarifications, though, if I may -

Not quite. The pitot system is sealed so that there is no flow through it. What it is doing is bringing the incoming airflow to a standstill which results in an increase in pressure. The pitot gives us static + dynamic pressure. By using the static input from the static port, we can get rid of the static pressure component and end up with dynamic pressure, from which we can figure out speed.

Again, not quite. The driver in calibration is the altimeter, for which there are error limits. Once these are met, the end result is that the ASI errors will be small and, sometimes, zero. All aircraft will have a PEC card, including Cessnas and Pipers. Because the PEC is small for the ASI (except around the stall), we usually can approximate CAS to IAS without too much problem.

That is to say, corrected for pressure/temperature, or density.]]>

The only practical comment to emphasise, is that, as part of the intercept of the final, desired track, you

(a) apply a suitable amount of lead to commence the turn (so you don't fly through the outbound track), and then

(b) roll out onto the expected heading, and then, for a little while

(c) check that the expected heading actually provides the correct navigation solution and you end up tracking the desired outbound track.]]>

]]>

On the right side of diagram there are 3 text boxes:

- Altitude with arrow pointing down,

- Transition Layer (shaded in the middle),

- Level with arrow up.

The correct answer is prior to >Layer<.

Levels are fixed, whilst Layer vary depending on qnh.

The X cross tells when to change, however there is a keyword "prior" in upper right corner.

Hope this helps]]>

In my book that has been amended I'll check the errata

Cheers]]>

If you were at 2500 assuming flying West bound and Restricted area goes from SFC to 9000, I would assume you would want to climb to first even cruising level above RA. You can't fly at 10500 which is transition layer.

Depending on qnh (hopefully given) you would fly at fl125 or fl145.

I would also expect distance from Restricted area was given too.

enr 1.7-9

Nav workbook page 55 para. 14.1.2.]]>

John]]>

I’ve started nav… but my text book is a 2009 edition… and already I can see some differences…. Like the 10% variable … still referred to as 15%… should I purchases a newer edition… as all of my exercises are also going to be out .]]>

Nav workbook , ta appreciate the help brack down on this quarter deal .

Does this mean the reference of the flight is changing ? ]]>

It is not easy to offer comment as to how long is a bit of string if we can't see the bit of string.]]>

All of my calculations and answers excluding the first one which is correct, differ quite a lot from actual answers.

Am I the only one with this problem? or putting it as politely as possible is the whole book riddled with errors here and there?

Also answers that should be rounded up i.e 31.765 gallons are consistently rounded down to 31.7 instead of 31.8 & vice versa it goes on, and it's always conflicting due to these inconsistencies, which i find quite frustrating.]]>

- The Bob Tait Navigation Text Book and Navigation Workbook were perfect preparation for the exam! The 1:60’s in the book are so helpful! I did those about three times leading up to the exam!

- I highly recommend PPE for Vertical Nav questions as I had 2 identical haha! The questions in PPE for RoC and RoD are harder than what I came across on the actual exam! Bob Tait also very helpful!

- I only got four 1:60’s. Two of which required me to determine the Track Error, one asked for drift experienced and the other one was a heading change required. Only converging and diverging 1:60’s. No cross track for me!

- Fuel calculations were also relatively basic. The textbook has everything you need. I got asked 2 questions on endurance remaining and 1 on finding the fuel flow.

- The first two questions were finding HW and XW! Make sure to use your ERSA for X-Wind calculations to ensure that you have the correct value. I know some people who used their flight computer only and were off by 2-3 knots! They are box answers as well! A few heading and ground speed questions too!

- Note that there are many box answers in the Nav exam! So make sure you take your time with every question and try to be as accurate as possible!

- In terms of the theory side of the exam I would make sure that you’re all over the first few pages of the Nav book! The questions are quite difficult to break down as they are worded typical CASA style! But if you have a sound understanding of the form of the earth, true + magnetic direction, chart projections + scale, and can wrap your head around the questions in Nav Progress Test 1 then you’ll be fine! I had about eight 1 markers on this chapter

- I did not focus enough time on the first chapter and therefore was not up to standard for those questions in the exam. That’s where all my KDRs were! Lat and Longs!

- No GNSS questions whatsoever!

- My number 1 tip is to TAKE YOUR TIME! I worked from the top to bottom and went thru slowly but swiftly. I had around 30 minutes to go but didn’t submit my exam till there was 11 minutes left. I didn’t change any of my questions on fuel and vertical nav as I was sure they were correct the first time. But yeah, take your time! Good luck to those sitting in the future! You’ll all smash it!]]>

Bob]]>

That's a great result and thanks for the feedback it's always appreciated

Cheers]]>

(b) simple enough standard use of any of the navigation calculators (or, if you really want to be fancy, you can do it totally trigonometrically with the CR as you have the trig tables on the slides)

(c) you need to be just as competent in this one as any of the standard calculations.]]>

Hope this is helpful ]]>

(a) we are looking at a single leg. First assumption is to see if we can run the calculations with a constant wind.

(b) G/S = 117 kts. That's just too close to 120 kts to be coincidence (2 nm/min).

(c) planned interval to the first fix is made good. G/S matches plan.

(d) time expected at the next fix comes up 4 nm short of the fix so the wind has changed and G/S has reduced. Therefore, the time to get to the second fix will be a bit more than 2 min, ie we will be at least 2 min late at the second fix.

(e) distance to the destination is a bit less that from the first to second fix so, presuming that the wind change holds for the last calculation, we should be about 2+ mins (for the lost time from the first to the second fix) + 2+ mins (for the second fix to the destination) which is going to be, say, around 5 minutes late at the destination (give or take a bit).

The answer is going to be (d).]]>

I sat NAV CPL a year ago

I'm now doing my KDR, on obstacle clearance. Iv forgot the equation on how to calculate the clearance of a obstacle, from a certain distance from the threshold

Can you please screenshot one of your equations

Regards]]>

GS = 133 kts

ETA at destination = 0304

hindi.fullform.website/exams/upsc/]]>

Solid result and thanks for the feedback

Cheers]]>

I managed to source a loan for the exam, perhaps I will invest in a new one also.

Cheers]]>

Just a problem that, sometimes, folks make the exam type questions too complicated and it all turns upside down ...]]>

Here, here!]]>

I totally agree with you John, Bob and Stuart do a great job keeping up with the constant changes.

Thank you kindly.]]>

For M 030 060

Steer C 028 060

The compass heading is 2 under for 030 and is correct for 060. So halfway between it would be 1 under. Nobody is interested in fractions of degrees on a compass, so if the magnetic heading is closer to 030 than it is to halfway, use 2 under, if it is closer to half way than to either 030 or 060, use 1 under and if it is closer to 060 than halfway, make no allowance.]]>

Also, these days, the question usually requires you to type in an answer on the computer. In this case CASA has indicated that you should type in the nearest whole number (no decimal places). The marking scheme will allow a small margin of error in most cases.]]>

As the earth rotates, all places on the same latitude will experience the same duration of daylight.]]>

thanks for the fast reply.

I mean just for the tabulation to litres, 1 USG= 3.785L, do we need to multiply by specific gravity?

In the question above, the calculation is 57 USG = 1 USG/ 3.785L = 57 * 3.785 = 215.745L

thanks.

on a note. i have keep typing the wrong password for samdol1978 user account. My IP is 118.189.185.198. Please assist to whitelist.

Thanks]]>

Able to share what was frequently touch on for your paper?

BOD, EOD, PNR, fuel margin, TMG/HDG and FPT? Anything else amiss?]]>

I find the Workbook a little confusing - because are you meant to work through the Topics in the main book and then go to the Workbook?]]>

You are correct those answers are for a different questions strange that has never been pointed out before I'll amend the text to reflect that

Cheers]]>

The picture I attached shows my way to find the answer.

I can't find anything wrong with my method.

I think the main problem is that the TE which is 44 degree is not applied to 1 in 60 thumb of rule as normally the TE should be less than

30 degree.]]>

This part made it click:

(1) Referring to this question in Nav practice test 1:

The position refers to crossing the railway line 35nm SE of YSGE b/w Bonathorne and Noondoo township. However, this point is in fact SW of YSGE.

(2) As mentioned by an earlier post, some of the practice questions refer to the Bourke WAC, and many students may not have purchased this given it isn't a requirement for the exam. Would it be worth changing these questions to refer to routes on the Sydney WAC?]]>

Cheers.

Lucas.]]>

Did you end up finding out where to find the GAFs needed to travel across a few states?

Cheers.

Lucas.]]>

The longitude difference between Perth and Canberra is 34°. So Canberra is 34 ÷ 15 = 2.26 hours ahead of Perth. That's 2 hours 16 minutes (to the nearest minute).]]>

That is really up to you, the workbook is a exercise book as in lots of repetition of exercises to help make you proficient, The exam preps are different as it's trying to give you a feel of how the CASA exams are presented. The exam preps would be my suggestion along with the workbook

but it depends on how you confident you feel your knowledge of the subject is.

Cheers

Stuart]]>

(a) 60nm

(b) less than 60nm

(c) greater than 60nm

In that case the answer would be 'less than 60nm' because the meridians converge towards the pole, one degree of longitude is not a constant distance. The distance between meridians gets less and less as you move towards the pole.

Way back in the 70's when I did my SCPL NAV (now ATPL) the syllabus covered a topic on Departure - the distance between to points on the same parallel.

In short, the formula for departure (in nm) was relatively simple:

In the example above, the distance (departure) between Town A and Town B can be calculated as follows:

Dep = ChLon x Cos(Lat)

Dep = 60 Cos 30 = 51.9 nm

If the towns were on the equator the distance between them would be: 60 Cos 0 = 60 nm

If the towns were at 90S the distance between them would be: 60 Cos 90 = 0 nm (meridians converge at the pole)

Apart from the SCPL NAV exam I don't recall ever using departure to calculate the distance between two points!]]>

Very cogent point.

However,

(a) if one has a debimeter or similar, one is in front

(b) even an accurately known (ie generally dipped) pre-start quantity plus known consumption rates against the clock and the flight log, with a gross check against contents gauges ... will provide a reasonable starting point.

Much easier on big iron ..

Harold's story makes for riveting reading.

Another case I recall involved one of the airline's flight standards bosses in the left seat .. PER-MEL, as I recall .. didn't end up too promising at MEL .. eventually a very close unplanned diversion to CBR and the noises started stopping on the taxy in to the terminal.

A more recent 737 case involving two aircraft can be reviewed here - www.atsb.gov.au/publications/investigati...13/aair/ao-2013-100/.

Historically, Australia is a right mongrel for unforecast fog, especially during the winter months. I have never been caught out to the degree of the cited stories but, on a number of occasions, we had sweaty palms running and re-running the fuel calculations in flight.]]>

I see where the issue is. You lose the half hour in the UTC to LST conversion. I got the same times as yourself give or take a minute with those AIP graphs.

Moral of the story. Remain in UTC time.

Thanks for your help Dav

Dobbo]]>

Thank you for your explanation, that is the answer i got, I must have read the question wrong somewhere...

Appreciate your help.

Cheers Dan.]]>

This is where I’m a little confused, I understand the first question which relates to the QNH being 1004 which makes cruising at 11,000 not allowed but when the QNH is more than 1013 (1028) wouldn’t the lowest fl available be 11,000?.... the transition attitude is always 10,000 so that’s 1,000 separation...]]>

AIP states 'The system of altimetry used in Australia makes use of a

transition layer between the Transition Altitude which is always

10,000FT and the Transition Level of FL110 to FL125 depending

on QNH (see Figure 1)'

My tired eyes read 'or' rather than 'to' between FL110 and FL125. Small text!]]>

Thinking back to the high school science classroom, the subject of measurement (fuel dip stick) and arithmetic based on those measurements may come to mind. More specifically, there are rules for determining the number of significant figures in quantities calculated from measured quantities like fuel.

Within the scope of this original post, say we dip the tank and get a reading of 85 USG - this is a measurement to 2 sig. figs. Now if we wish to convert to litres we can use any of the conversion factors (3.7854, 3.78, 3.79, 3.8) discussed earlier and get the same answer to 2 sig. figs.

85x3.7854 = 321.759 = 320 to 2 sig. fig.

85x3.78 = 321.3 = 320 to 2 sig fig

85x3.79 = 322.15 = 320 to 2 sig fig

85x3.8 = 323 = 320 to 2 sig. fig.

(Note that the red 0 in 320 is not significant. If it were the answer would be 320. with a decimal point after the 0)

In summary, we can use as

I don't know how CASA would deal with an answer like 320, but 320 lts is the mathematically correct conversion.

See Wikipedia for a more detailed discussion on significant figures

I have completely ignored the subject of the propagation of errors.]]>

Cheers

Stuart]]>